Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace transform of dirac delta function

  1. Nov 14, 2004 #1
    let S be the Unit Step function

    for a function with a finite jump at t0 we have:

    (*) L{F'(t)}=s f(s)-F(0)-[F(t0+0)-F(t0-0)]*exp(-s t0)]


    L{S'(t-k)}=s exp(-s k)/s-0-[1-0]*exp(-s k) = 0 & k>0

    but S'(t-k)=deltadirac(t-k) and we know that L{deltadirac(t-k)}=exp(-s k)

    so why do I get ZERO when using the formula (*)
  2. jcsd
  3. Nov 15, 2004 #2


    User Avatar
    Science Advisor

    Ok, I'm not familiar with that one, but I think you're mis-applying it.

    The expression I'm familar with is just the L{F'(t)}=s f(s)-F(0) part.

    My suspicion is that the [F(t0+0)-F(t0-0)]*exp(-s t0)] is just inserted to manually take care of the dirac impulse that results from the finite discontinuity and that in this case the f(s) you should be using is that of the original function without the discontinuity. Note that the unit step without the step is a pretty simple function. :)
    Last edited: Nov 15, 2004
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook